\(a,NTK_X=\dfrac{5,31.10^{-23}}{0,16605.10^{-23}}=32\left(đ.v.C\right)\\ \Rightarrow X:Lưu.huỳnh\left(S=32\right)\\ b,NTK_X=\dfrac{4,482.10^{-23}}{0,16605.10^{-23}}=27\left(đ.v.C\right)\\ \Rightarrow X:Nhôm\left(Al=27\right)\\ c,NTK_X=\dfrac{10,6272.10^{-23}}{4.0,16605.10^{-23}}=16\left(đ.v.C\right)\\ \Rightarrow X:Oxi\left(O=16\right)\\ d,NTK_X=\dfrac{33,21.10^{-23}}{5.0,16605.10^{-23}}=40\left(đ.v.C\right)\\ \Rightarrow X:Canxi\left(Ca=40\right)\\ e,NTK_X=\dfrac{23,5791.10^{-23}}{4.0,16605.10^{-23}}=35,5\left(đ.v.C\right)\\ \Rightarrow X:Clo\left(Cl=35,5\right)\)
a) \(NTK_X=\dfrac{5,31.10^{-23}}{1,6605.10^{-24}}\approx32\left(đvC\right)\)
`=> X: S` (lưu huỳnh)
b) \(NTK_X=\dfrac{4,482.10^{-23}}{1,6605.10^{-24}}\approx27\left(đvC\right)\)
`=> X: Al` (nhôm)
c) \(NTK_X=\dfrac{10,6272.10^{-23}}{4.1,6605.10^{-24}}=16\left(đvC\right)\)
`=> X: O` (oxi)
d) \(NTK_X=\dfrac{33,21.10^{-23}}{5.1,6605.10^{-24}}=40\left(đvC\right)\)
`=> X: Ca` (canxi)
e) \(NTK_X=\dfrac{23,5791.10^{-23}}{4.1,6605.10^{-24}}=35,5\left(đvC\right)\)
`=> X: Cl` (Clo)
